Graphs of standard functions

Correct answer: A. (x=-2) and (y=3)

The denominator (x+2) is zero at (x=-2), and the outside (3) gives horizontal asymptote (y=3). In exams, check both denominator and vertical shift.

Correct answer: A. ( -6 )

For (x\ge5), (|x-5|=x-5) and (|x+1|=x+1), so (y=-6). In exams, open modulus according to the interval.

Correct answer: B. ( -2 )

When (x\in[-3,-2)), (x+1\in[-2,-1)), so (\lfloor x+1\rfloor=-2). In exams, first transform the inside interval.

Correct answer: A. ( -1 )

The direct answer is option A, -1. The signum function returns 1 when its input is positive, 0 when its input is zero, and -1 when its input is negative. If x>2, subtracting x from 2 gives 2-x<0. Therefore sgn(2-x)=-1, so the graph value is -1. Option A is correct. Option B, 0, would occur only when 2-x=0, which means x=2, not x>2. Option C, 1, would occur when 2-x>0, meaning x<2. Option D, 2, is the constant appearing inside the expression, not a possible signum output in this situation; signum outputs only -1, 0, or 1. The boundary x=2 is important: the value there is 0, while all values strictly to the right give -1. The supplied answer is consistent. Remember to determine the sign of the entire inside expression before applying signum.

Correct answer: A. \(x=-5\) and \(x=1\)

Every point on the \(x\)-axis has \(y=0\). Hence, \(3|x+2|-9=0\), which gives \(|x+2|=3\). Taking both cases, \(x+2=3\) gives \(x=1\), while \(x+2=-3\) gives \(x=-5\). Therefore, the intercepts occur at \(x=-5\) and \(x=1\). Option B does not result from solving the modulus equation correctly. Exam tip: for \(|u|=a\) with \(a>0\), always consider both equations \(u=a\) and \(u=-a\).

Correct answer: A. ((-\infty,5])

Since (\sqrt{x+4}\ge0), (5-\sqrt{x+4}\le5). In exams, a negative square-root term makes the graph go downward.

Correct answer: B. (3) left, vertical stretch by (2), (1) down

(x+3) shifts (3) units left, outside (2) vertically stretches, and (-1) shifts down. In exams, read inside and outside transformations separately.

Correct answer: A. (x=-1) and (x=3)

Equating gives (x^2-1=2x+2), that is (x^2-2x-3=0). In exams, set the two (y)-values equal for intersections.

Correct answer: A. (x=4)

The denominator (x-4) becomes zero at (x=4). In exams, find the vertical asymptote of a reciprocal graph from the denominator.

Correct answer: A. ((4,0))

From (3x-12=0), (x=4) and the minimum is (0). In exams, find the vertex by setting the inside of the modulus to (0).

Correct answer: A. \(3\) and \(4\)

In this interval, \(2x\) starts in ([3,4)) and also reaches values giving \(4\) after \(2x\ge4\). In exams, split the inside interval into subintervals.

Correct answer: A. circle with centre ((0,0)) and radius (4)

Squaring gives (x^2+y^2=16) with (y\ge0). In exams, identify (\sqrt{r^2-x^2}) as an upper semicircle.

Correct answer: A. \(f(x)=\operatorname{sgn}(x)\)

For \(x>0\), \(|x|=x\), so \(\frac{|x|}{x}=1\). For \(x<0\), \(|x|=-x\), so \(\frac{|x|}{x}=-1\). Therefore, its graph is the same as \(y=\operatorname{sgn}(x)\) on the domain \(x\ne0\); the function is not defined at \(x=0\). Option B reverses both values, while options C and D vary continuously with \(x\). Exam tip: split expressions containing \(|x|\) into the cases \(x>0\) and \(x<0\).

Correct answer: A. ((2,4))

The minimum of (\sqrt{x-2}) is (0) at (x=2), so (y=4). In exams, a negative square-root graph has maximum at the starting point.

Correct answer: A. (x=3)

The axis of symmetry is (x=-\frac{b}{2a}=-\frac{-12}{4}=3). In exams, use the axis formula for a quadratic graph.

Correct answer: A. (y=\frac{5}{2})

The degrees of numerator and denominator are equal, so the ratio of leading coefficients is (\frac{5}{2}). In exams, use the ratio of leading coefficients for equal degrees.

Correct answer: D. ((3.8,0))

(\lfloor3.8\rfloor-2=1), so ((3.8,0)) is not on the graph. In exams, first find the greatest integer of (x).

Correct answer: B. (5)

The distance between (7) and (2) is (5), so the minimum sum is (5). In exams, remember the minimum of (|x-a|+|x-b|) is (|a-b|).

Correct answer: A. maximum (9) when (x=2)

The minimum of (|x-2|) is (0) at (x=2), so (y=9) is maximum. In exams, a modulus graph with a negative multiplier has maximum at the vertex.

Correct answer: A. \(y=x^2+2\)

For \(y=x^2+2\), \(f(-x)=(-x)^2+2=x^2+2=f(x)\), so it is an even function and its graph is symmetric about the y-axis. Since \(x^2\geq 0\), we have \(y=x^2+2\geq 2\); hence the minimum value is 2, attained at \(x=0\). Option C has minimum value 0 and is symmetric about \(x=2\), not the y-axis. Exam tip: test y-axis symmetry using \(f(-x)=f(x)\).

Correct answer: A. (\mathbb{R}\setminus{1})

The denominator (|x-1|) becomes zero at (x=1), so (1) is excluded. In exams, remove values that make the denominator zero.

Correct answer: A. ( -1 )

(\lfloor1.7\rfloor=1) and (\lfloor-1.7\rfloor=-2), so the value is (-1). In exams, avoid mistakes with greatest integers of negative numbers.

Correct answer: D. ( 2 )

For (1<x<4), (\operatorname{sgn}(x-1)=1) and (\operatorname{sgn}(x-4)=-1). In exams, check the sign of each signum term separately.

Correct answer: A. (x=1) and (x=3)

The denominator (x^2-4x+3=(x-1)(x-3)) is zero at (x=1) and (x=3). In exams, find vertical asymptotes from the zeroes of the denominator.

Correct answer: C. ( 3 )

The direct answer is option C: the constant value is 3. When x\ge4, both x-1 and x-4 are non-negative, so |x-1|=x-1 and |x-4|=x-4. Substitute these forms: y=(x-1)-(x-4)=x-1-x+4=3. Therefore the expression is constant throughout the interval x\ge4. Option C is correct. Option A has the wrong sign. Option B would mean the two absolute values cancel completely, but their difference is 3. Option D, 2x-5, results from an incorrect sign expansion and is not constant. Memory cue: first identify the interval, then remove each modulus with the correct sign.